A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work?

A's 1 day work = $\dfrac{1}{20}$

B's 1 day work = $\dfrac{1}{15}$

(A + B)'s 1 day work = $\dfrac{1}{20} + \dfrac{1}{15}$

= $\dfrac{3 + 4}{60}$

= $\dfrac{7}{60}$

(A + B)'s 6 day work = $\dfrac{7}{60} \times 6$

= $\dfrac{7}{10}$

Remaining work = $1 - \dfrac{7}{10}$

= $\dfrac{10}{10} - \dfrac{7}{10}$

= $\dfrac{3}{10}$

No. of days taken by B to finish the remaining work = $\dfrac{\text{Remaining work}}{\text{B's 1 day work}} = \dfrac{\dfrac{3}{10}}{\dfrac{1}{15}}$

= $\dfrac{3 \times 15}{10 \times 1}$

= $\dfrac{45}{10}$

= $\dfrac{9}{2}$

= $4\dfrac{1}{2}$

Hence, B will take $4\dfrac{1}{2}$ days to finish the remaining work.